Solve the equation $\;$ $x^4 - 8 x^3 + 24 x^2 - 32 x + 20 = 0$ $\;$ if $\;$ $3 + i$ $\;$ is a root.
Given: $\;\;$ One root is $\left(3 + i\right)$
$\implies$ $\left(3 - i\right)$ is also a root.
Sum of roots $= 6$
Product of roots $= \left(3 + i\right) \left(3 - i\right) = 9 - i^2 = 10$
$\therefore$ $\;$ The corresponding factor is $\;$ $x^2 - 6x + 10$
$\therefore$ $\;$ $x^4 - 8 x^3 + 24 x^2 - 32 x + 20 = \left(x^2 - 6x + 10\right) \left(x^2 + \lambda x + 2\right)$ $\;\;\; \cdots \; (1)$
$\lambda$ $\;$ is a real number (constant).
Equating the coefficients of the $x$ term in equation $(1)$ we have,
$- 32 = -12 + 10 \lambda$
i.e. $\;$ $10 \lambda = -20$ $\implies$ $\lambda = -2$
Substituting the value of $\lambda$ in equation $(1)$ we have,
$x^4 - 8 x^3 + 24 x^2 - 32 x + 20 = \left(x^2 - 6x + 10\right) \left(x^2 - 2x + 2\right)$
$\therefore$ $\;$ $x^4 - 8 x^3 + 24 x^2 - 32 x + 20 = 0$ $\implies$ $x^2 - 6x + 10 = 0$ $\;$ or $\;$ $x^2 - 2x + 2 = 0$
Now,
$\begin{aligned}
x^2 - 2x + 2 = 0 \implies x & = \dfrac{2 \pm \sqrt{4 - 8}}{2} \\\\
& = \dfrac{2 \pm 2i}{2} \\\\
& = 1 \pm i
\end{aligned}$
$\therefore$ $\;$ The roots are $\left(3 \pm i\right)$ and $\left(1 \pm i\right)$.